Upper and Lower Bounds for First Order Expressibility

Neil Immennan

Computer Science DepartmentCornell UniversityIthaca, New York 14853

OVar &Sz[v(n),z(n)] = ASPACE &TIME[v(n),t(n)]

That is variables and size correspond preciscly to alternating spaceand tiane respcctively. Returning to variablcs ranging ovcr an ne)enlcnt universc, it fo)Jows that:

Abstract

We continue the study of first order expressibility as a measureof complexity, introducing the new class Var &S7~v(n),z(n») oflanguages expressible with v(n) variables in sentences of size z(n).We show that when the variables are restricted to boolean values:

Var[O(l)] ASPACE[log n] PTIME

and noticed that only a constant nUlnber of variables were needed.Furthennore while the existential quantifiers range over theelemente; of the universe of the input. (Le. 1 to n), the universalquantifiers could be boolean. Thus we let Var &SzIv(n),z(n}] be theclass of properties unifolmly expressible with exactly v(n) variablesand size Olz(n)]. Also let BUVar &Sz[v(n),z(n)] be the saIne classwith the additional restriction that the universal quantifiers areboolean. Let "*" abbreviate 0[1]. We show that:

NSPACF~logn] ~ BUVar &S7~~,lognl ~

~ Var&Sz[*,logn] ~ DSPACE[log2(n)]

That is the family of properties uniformly expressible with aconstant number of variables is just PTIME.

These results hold for languages with an ordering on the objects inquestion, e.g. for graphs a successor relation on the vertices. Weintroduce an "alternating pebbling game" to prove lower bounds onthe number of variables and size needed to express propertieswithout successor. We show, for example, that k vatiables areneeded to express Clique(k), suggesting that this problem requiresDTIMF~nk] .

Introduction nod Sunnnary

In [fmm79] we proposed studying the complexity of a property,C, via the size of a sentence from first order logic n~eded to expressC. We showed there that the memory space needed to check if agiven input has property C is closely related to the size of C'sS1l1al1est first order description. More precisely:

NSPACElftn)] ~ Size(tln)2 /Iog(n)] ~ DSPACE(t{n)2]

Here Size[g(n») is the family of all properties expressible by auniform sequence of sentences, F1 F2 ... , where Fn has O[g(n)]symbols.

I~eading some papers by RUZlO ([Ruz79a), [Ruz79b», onsimultaneous resource bounds, we tried to find anaJagous results forfirst order expressibility. First we reexamined our proof of theabove containtnent for f(n) =]og(n), i.e:

NSPACF~lognJ ~ Size[logn] ~ DSPACF~log2(n)],

* Research partly supported by NSF grant no. MCS 78-00418

CH1498-5/80!0000-0074$OO.75 © 1980 IEEE74

Savitch's simulation of NSPACE[logn] by })SPACE[log2(n)] maybe optimal, but one way of thinking of the difference bctween theclasses is that DSPACE[log2(n)] can simulate log(n) universalquantifiers ranging from 1 to n. We conjecture that all threecontainments in the above chain are proper, but none are known tobe.

It turns out that BUVar&Sz(*,log(n}] is idcntical to the naturalclass Log(CFL) -- those languages log-space reducible to somecontext free language. We will also see that the third telm in theabove chain, Var&Sz[*,log(n)1 is equal toASPACE&Alt[log(n),1og(n») -- the class of languages accepted by anASPACE[log(n)] Turing machine which Inakes only O[log(n)]alternations between existential and universal states.

Once the idea of counting distinct variables was raised it wasnatural to relax the size restriction. J)efine Var[*] = Uk =1.2... Var

&SzIk.nk] -- those properties expressible with a constant number ofvariables. It turns out that Var[*] is identical to polynomial time!The identity between P and Var[*] is a very pleasing result bothbecause it indicates that first order expressihility is a fruitful view ofcOlnplexity, and because it is another delnonstration of thefundaiTIental importance and model independent nature of P.

One weakness of our previous definition of exprcssibility size isthat it makes usc of the notion of Turing machines in the definitionof a "unifonn" sequence of sentences. Our feeling at the time wasthat the uniformity condition was an imperfect attempt to capturethe notion that we rcaJly had one sentence with a variablc numberof quantifiers, just as we have the notion of one Turing machinewith a variable amount of space. Indeed the use of constantlymany variables leads us to the realization that ther~ is a syntacticuniformity -- the nth sentence ofa Var &Sz(k,z(n)] property is justzen) repetitions of a fixed block of k quantifiers. rInis newdefinition of uniformity makes Var[*] entirely a notion from logicand thus increases the interest of the fact that it is equal to P.

Section A: J)efinitions and Motivations

'Ix 3y [ E(x.y) or F~y,x) ]

Continuing in our rcfinement notice that when wc writePn/2(u,v) we may reuse x,y,z -- their current values are no longerneedf'o. Being slightly wasteful for the sake of clarity. write:

Pn(X,y) - 3zVuVw (lu=x&v=z or u=z&v=y] =>

3x3y[x= u&y =v & Pn/2(X,y») (4)

(1)

(2)

{ G I a-+*-+b }GAP

P1(X,y) == (x = y) or E(x,y)

Pn(x,y) == 3~ ( pn/2(x,z) &Pn/2(z,y) )

Think of a directed graph, for example, as a universe, V, thevcrtices, together with a binary edge relation E(-,-) on V. This is alogical structure of silnitarity~ Ta = {E(-.-)}. The language ofa type, T. L[T]. consists of the sentences built up frOln the symbolsof T using the logical connectives &. "or", I, =>, variables x,y, ... ,=, and quantifiers, 3x and 'Ix. ranging over the universe. Forexample, considcr the following sentence from L[Toj:

GAP is known to bc complctc for NSPACE[log nl. (See [Sav73].)We show in [lm80a] that GAP is complete in a very strong scnse -­every problem C in NSPACE[log oj has a first order sentencetranslating all instances of C into instances of GAP.

To express GAP we will write. down fOlmulas Pn(a,b) meaning,wlberc is a path of length at most n from a to b." Wc dcfine Pnby induction as follows:

S1 says that cach vertex, x, has an cdge cOIning out of it or an edgegoing into it. A graph satisfies S) (in symbols OI=Sl) if it has noisolated vertices. Notc that every graph 0 "understands" everysentence S from I.[Ta], Le. Gt=S or GI=IS.

To motivate thc definitions for variable and sizc expressibility wenow consider a stepwise refinement of sentenccs exprcssing aspecific problem. Let GAP be the set of directed graphsG withsped fied points a and b such that there is a path in G from a to b.In symbols:

We propose to study the complexity of a condition, C, by asking,"How difficult is it to express C?" For this expression we choosethe natural first order language of the objects undcr consideration.

Equation (2) defines Pn in a way that increases the quantifierdepth by one each time n is doublcd. However Pn/2 is writtentwice on the right so the size of this Pn is twice the size of Pn/2 •We can alleviate this problem using' the "abbreviation trick" (seec.g. [FiRa74]). The trick uses uni"crsal quantifiers to pcnnit us towrite Pn/2 only once on the right. Thus:

Pn(x,y) == 3zVuVw [u=x&v=z or u=z&v=y ] => Pn/2(u,v) (3)

We have now written Pn with O(1og n] symbols, thus provingthat GAP is in Size[log n], to be defined.

The gatnes mentioned above give us lower bounds only on whatcan be expressed without the successor predicate. We show, forexmnplc. that Clique(k) -- the existence of a cOlnplete subgraph onk vertices -- cannot be expressed with k-l variables. without Sue.(Of course k variables suffice -- just say there exist xl ... xk forminga clique.) This is a plausability argument that Clique(k) is not inVar[k-l]. If we cpuld prove the latter result, i.e. that Clique(k)cannot be expressed with k-l variables in the language with Suc,then it would follow that the genel:al clique problem is not inVar[*)' From this it would follow that p;t= NP.

Our definition of Val' &Sz gives the sentences access to somearbitrary successor relation. Suc(-.-), on the universe of the inputstructures. Without this added relation we cannot silTIulate Turingnlachines -- there is no way to say, "Now the Turing Inachinenloves its input head one space to the right." We, showed in[ltnm79] that Suc(-,-) is not needed to express certain "natural"graph problems such as connectivity; however, it is essential forother uses such as counting the parity of a totally disconnectedgraph.

Now that wc know that 1)'1'11\1 E[nk] is closely related to Var[k] it

is uscful to determine which graph properties can and· <;anot bcexpressed with k variables. In Section C we describe acOJnbinatorial gmne. a Inodification of Ehrenfeucht-Fraissc ganlcs,(see lEhr61] or [Fra54]). with which we can prove lower .bounds onwhat can be expressed in k variables. These new ganles are analternating version of pebbling games.

In Section D we also consider the graph isolnorphism problem.If we knew. for example. that Graphlso were in IJrIME[n1og(n)]then it would follow that this property could be written with log(n)variables without successor. Thus a pair of graphs, <G,H>,satisfyFn . a sentence with log(n) variables. if and only if they arcisolnorphic. Clearly <O.G>t= Fn . Now suppose that G==Var[JognlH,Le. G and H agree on all log(n) variable sentences. It follows that<G.H>t==Fn, and thus 0 and H are isomorphic. We have shown:

Oraphlso E Var(w.o. Suc)[v(n)] .....

VO VH (IOI=II-II=n => [G~H ++ G ==Var[v(n)] H])

In DTIME[n4v] we can check if G ==Var[v] H. Thus a nearoptitnal algorithm fi)r Graphlso may be to find the correct v(n) inthe above equivalence, and check if G ==Var[v(n)] H .

We mentioned above a proof that Clique(k) cannot be expressedin )(-1 variables without Suc. This is shown by building graphs Gand H' such that G .== Var[k-l] H and yet G has a k-cJiquc while Hdoes not. Clearly G and H are not isomorphic. Thus we have a kvariable (without Suc) lower bound on Graphlso. This is apJausahiJity argumcnt that' Graphlso is not in P.

In the following pages we give: .(A) Definitions and motIvations;(B) Statements of some of the main relati()nships bctweencxpressibility and Turing machine Time and Space; (C) Thealtermtting pcbbling game; (I) Probabilistic graph argumentsfollowing [Fag76] and [BIHa79] showing that Hamilton Circuit,Clique, and Graphlso are not in Var(w.o. Suc)[*]; and (E)Conclusions and directions for future rcsearch.

75

We have succeeded in expressing GAP by a uniform sequenceof sentences, {Pn(a,b) I n~ l}, such that Pn has 5 variables and sizeOlIogn]. This suggests the following:

Definition: A set C of structures of typc -r is expressible ill .YU!lvariables and size 1.!!}, (in symbols, C is in Var &Sl'!v(n),z(n»), ifthere exists '3 unifonn sequcnce of sentences F] F2 ... froJn,1.[-rU{Suc}) such that· :

a. For all structures G of type -r with IGI~ n, and for allSuco(·'·) a valid successor relation on the universe of a,

a f. C ++ <G,Suco>t= Fn

b. rn has v(n) distinct variables and a total of O[z(n)] sylnbols.

As Ruzzo has shown in [Ru79b], unifonnity conditions may begreatly varied without significantly changing a definition. Thefollowing condition will suffice in what follows:

Let's return to Equation (4) .and notice lhat in sinllllaling anNSPACE[logn] property, two universal quanlifiers ranging frOln 1to 11 are lIsed. Their purpose is only to m.ake a choice between thefirst half and the second half of the path. It nlakcs sense toInininlize the universal choices when sinllilating an existential classso we replace "VuVv" in Equation (4) by "Vb", where b is boole""valued. 'rhus:

Pn(X,y) == 3zVb3u3v ([(b=O&u=x&v=l) or (b= I &ti=l&V=Y»)

& 3x3y [x=u & y=v & Pn/2(X,y)]) (6)

l)Cfine BUVar&S:4v(n),z(n)] to be the farnily of propertiesexpressible in v(n) variables and sile O(z(n)J where the existentialquantifiers still range froln 1 to n, but the universal quantifiers arcboolean. Thus it is easy to. sec that GAP is in BUVar &Sz[k,logn],and Inore generally,

Uniformity Condition (*): The map n -+ Fn is generable inI)SPAC"~v(n).logn] and l)rrIM~lz(n)].

lllcorem A.2: For s(n) ~ 10g(n),

NSPAcrts(n)] ~ nUVar &Sz[O(s(n)/log(n») ,s(n) 2/10g(n)]

Of course (*) does not capture our intuitive feeling that the Fn'sare all the sanle sentence with varying numbers of quantifiers. Tomake the latter notion more precise abbreviate quantifiers withrestricted dOlnains as follows:

Section II: Variables~ Sile versus Tilne ~ Space

Unifonnity Condition (**): lbere exist constnnt c and fonnulasA,n, and quantifier free fonnulas MI'" Mv all of which havevariables only xl'" Xv such that:

Fn - A «Ql xI . M1 ) ••• (Qy Xy. My) ) c.1~n) times D

Now we can write Equation (4) more compactly a~:

Pn(x,y) = 3zVu(Vv. M3 )3x(3y . Ms ) Pn/2 (x,Y) (5)

Here M3 = [u=x&v=z or u=z&v=y], ~nd M5=[x=u&y~vJ.

Equation (5) gives a model for the followmg totally syntacticaldefinition of uniformity for Var&S4v,z(n») :

We adopt (**) as our definition of unifonnity for Var&S7Iv,z(n)] when v is a constant, otherwise we use (*). It followsthat GAP is in Var &Sl!5,logn), NSPACF~logn) is. in Var&Sz[k,logn), and indeed:

Theorem A.1: For s(n) ~ 10g(n),

NSPAC~ls(n)] C Var &Sl[O[s(I1)/log(n)] , s(n) 2/log(n»)- ~ I)SPAC~ls(n) ~

proof sketch: The proof is nearly the same as for Theorem 2 in.[lmm79). We showed there that NSPACl-ls(n») ~ ·Size(s(n)2/log(n)]~ IlSPAC"1s(n)2]. That proof noted that a Turing machineinstantaneous description (Ill) o( si7.e s(n) could be coded inO[s(n)/log(n)] variables since the variables range over an n clementuniverse. rlbus using equation (3) we asserted the existence of acomputation path of length e5fD); O[s(n)] ID's were needed. For theproof of the first inclusion in Theorem A.l we use equation (4)instead. 'll1us only a constant number of ID'~, requiringO(s(n)/Iog(n)] variables, must be remembered at once. I

Recall a definition and result of Sudhorough [Sud78):

f)cfinition: AuxPDA(s(n),t(n») is the class of languages accepted bya two way nondeterministic push down automaton with auxilliarywork tape of size s(n), running in time ten).

Fact (Sudborough): AuxPDA[1og(n), n*J = Log(CFL).

rnlenrem n.l: nUVar&SlIO(s(n)l1og(n») ,z(n)]= ASPACJ..:&:TS[v(n)log(n), *z(n~

In [Ruz79a] Ruzzo defines an accepting cmnputation tree of ~n

alternating Turing machine M to be a tree whose root is a startingII) of M, whosc nodes are intennediate II )'s, and whose Icaves areall accepting configurations. ~:C1Ch universal node,· tI, has all itspossible next moves as offsiJring, while the existcntial nodes, e, leadto exactly one of e's possible next moves. We say that a languageC is in "SPACE &TS[s(n).z(n») if all members of C of silc n areacce(>ted in a computation tree using space sen) and tree size,(number of nodes), z(n). RUllO relates this new measure toauxiHiary pda's via,

Fact (Ruzzo): A~PACE &TS[s(n),z(n)*] = AuxPllA[s(n),z(n)*).

Notice that both the tree size mode) and the Auxl)I)A chargemuch .nore for universal nloves than for existential ones. 'Incfoll()wing theorem shows that we get the Sc1me classes in ourexpressibiJity measure by restricting all universal quantifiers to beb(x)lean. In a sense we charge .Iog(n) times as much for a universalchoice as for an existential.

proof: (~): Given an input structure G with n clement universewe can generate Fn ' the nth sentence in our unifonn sequence.

We must show that in ASPACE &TS[v(n)log(n), *7.(n») we can checkif GI=Fn .

read, "There exists xsuch that M.It

read, "For all x such that M."(3x . M) [ ] == 3z [M & ]('Ix. M)[ ] == Vx[M => ]

76

2. Conditions which can be expressed with v variables are justthose conditions which can be checkcd in DTIM~lnO[v] ]

The above corollary rounds out a pleasing relationship betweenexpressibiJity and computation. We have shown:

1. The size of a sentence needed to express condition C ispolynomia11y related to the amount of memory space neededto check if C holds for an input, and,

(b): By our uniformity condition the nth sentence of a Var[v(n)]property can be generated in DSPACF.(v(n)log(n)] and so it is ofsize at most nO[v(n)] Restricting the universal quantifiers to beboolean at worst increases the size by a factor of log(n). Thus:

Var[O[v(n)Jj = Var&Sz[O[v(n)],nO[V(n)]]= BUVar&SltO[v(n»),nO[v(n)] ]

The next theorcm generalizes the above results giving aremarkably close relationship between expressibi1ity and alternatingmachine conlplexity. Let nVar &Sz[v(n),z(n)] be those propertiesexpressible in a sentence with v(n) booleall variables and sizez(n).For each predicate symbol, E, we add the predicate symbols, E:t 'E2 ••• where En (b1 ••• blogn ' CI ••• clogn ) means E(b,c) where b hasbinary vertex number til." b10gn •

111eorem B.3: ASPACE &TIMF~s(n),t(n)] = BVar&Sz[O[s(n)],t(n)]

I

Using Theorcm B.l,

= ASPACE&TS[v(n)log(n),2 nO[v(n)]]= ASPAC~1v(n)log(n)]

= DTIME[nO[v(n)]]

(c): This is a special case of(b).

proof: (;J): Given an input structure 0 with n element universewe can generate Fn the nth clement in our uniform sequence. Wemust show that in ASPACE&TIMF~s(n),t(n)] we can check ifGt=Fn •

To test.if G satisfies Fn we read the sentence from left to rightholding the present values of the variables bI ... bs(n) in ourmcmory. At quantifiers 3bi or \fbi we make the appropriateexistential or universal choice of a new value for bi . Similarly at&'s (or "or"s) we universally (or ·existentially) choose one branchand proceed. We have G and the values of the variables so wemay check the truth of atomic formulas: b=0, or E(bl ... clogn ) , in

number of.steps a constant times their length.

( ~): Going the other way we must write the sentence,Acceptt(r) meaning that alternating Turing machine M when startedat instantaneous description r will reach acceptence in tsteps. Weaccomplish this by saying that if r is existential then there existssome next step x and Acceptt_l (x), whereas if r is univer&1l then forall next steps x, Accep~_l (x) .

A technical difficulty here is that if at each step wc recopy theentire 10 then the size of the resulting sentence will be s(n)t(n). Tosimplify the problem let us assume that the moves of our Turingmachine alternate at every step and branch into at moSt two moves.'Ibus we can write,

Accept(r.<ql ,Q2 ,Q3 ),z) - 3p ( Accept(r,<ql ,p),z) &

Accept(p,(q2,q3)'z) )

BUVar &Sz[*,log(n») ASPACE&TS[log(n),n*)= AuxPDA[log(n),n*)= Log(CFL)

3p (3s1 82 s3 ' a permutation of Ql q2 q3 )( Accept(r,(sl ,p>,z) &

Accept(p,(s2 ,s3 ),z) )

To test if G satisfies Fn we read the sentence from left to rightholding the present values of variables xl ... xv(n) in our v(n)log(n)memory. Note that each non-boolean variable may have value 1 ton corresponding to an clement of G. At existential quantifiers, 3xi,

we existentially choose some xi from the universe of G and at

universa.l choic~s, Vbj , we universally choose bj . When we cometo atomic predicates, e.g. F~xI ,x2 ) or bI7 =0, we can check theirtruth because we have the current values of the variables. Notethat this accepting procedure has tree size *z(n) because we maymake a bin,try universal split O[z(n)] times.

( ;J): Here we fonow a proof of RUllO [Ruz79a). We mustexpress the property Accept(r,z) which means that the alternatingTuring machine M will accept in tree size z when started with ID r.We express Accept(r,z) by choosing a point p in the middle of thetree whose subtree is of size between 1/3 and 2/3 of the originaltree. 'Ibus,

Acccpt(r,z) == 3p ( Accept(r,(p),(2/3)z) & Accept(p,<>,(2/3)z) )

Here Accept(r,(ql ... qt),z) means that there is a computationtree. of size z starting at r such that each leaf is either an acceptingconfiguration or one of qI ... qt .

Our only trouble is to insure that the list (ql ... qt> stays ofconstant size. Whenever the list is of length three we take an extramove to split it in half by finding a point p above two of the threenodes in the list,

Thus we can write Accept(..) with a constant number of ID's, Le.v(n) variables, and the size of the sentence is O[log(z)]. 1bisproves Theorem B.1. I

Note that in the above we can add a boolean universal quantifierand use the abbreviation trick to write Accept(-) only once on theright. Also note that the above is a slight lie since we don't knowwhich pair of q's p will be above. In fact we would have to say,

Corollary. B.2:

a. BUVar&Sz{*,log(n») = Log(CFL)

b. Var[O[v(n)H = DTIME{nO[v(n)] )

c. Var(*] = PrIME

proof:(a): From the above theorem, together with the results from

Ruzzo and from Sudborough:

77

Let ASPAC~.:&Alt(s(n),a(n)] be those problems accepted inalternating space s(n) with at most a(o) alternations betweenexistential and universal states. 'Illen:

Corollary B.4: For s(n) ~ 10g(n),

ASPACE&TIME(s(n),t(n)] ~ Var &8z(O[s(n)/10g(n)],t(n)]~ ASPACE&TIMF~s(n),t(o)log(n)]

Corollary U.S: Let s(o) ~ a(n)log(n). '!ben:

ASPACE.&Alt[s(n),a(n)]~ Var&Sz[O[s(n)/log(n)],(a(n)+s(n»s(n)/logn]

and in particular, ASPACE &Alt(log n,log n] =Var &Slt*,log n1

proof: To assert the acceptence of an ASPACE&AJt(s(n),a(n)]computation we assert the existcnceof a(n) ID's where thealternations occur. Each path hetwecn the IO's has no alternationsand so can be expressed in Var&S7JO[s(n)/log(n») ,. s(n)2 /Iog(n) ]by Theorem B.1. We write ACCEPTa (100 ) to mean that 100leads to acceptencc in a alternations:

Here "II)o-+bl ... bs-+ID1" means that there is a computation

whose jth movc makes the bj th choicc and leads from 100 to IDl ins stcps. lois is a detenn inistic computation of length sand,although we omit the details, it can be asserted to exist with O[s]symbols. I

\Ve have demonstrated an exact relationship betwecn alternatingTuring machines and quantified boolean formulas. If we return tothe more natural language of the input stl1Jctures, Le. variablesranging from 1 to 0, then the needed nUJnber of variables tositnulatc s(n) space becomes s(n)/log(n). It is not clear, however,how to do better than size ten) to sitnulate time t(n) because themachine might go through ten) alternations. Thus we can onlyshow:

Corollary 8.6: NSPACfllogn] ~ BUVar &Sz[k,logn]~ Var &S7Jk,logn] ~ OSPACF~log2 (n)]

Corollary 8.6 which comes immediately from Theorems B.l andB.2 sheds some light on the difference between DSPACE(log2 (n)]and NSPACE[logll]. We conjecture that all three containmentsabove are proper, but it is not even known that NSPACE[log(n»):I: DSPACF~log2(n)] .

It makes sense to consider a sentence with k variables which isof length greater than nk • Similarly we can consider anASPACFtlogn] machine whieh runs for more than nk steps.Generalize the definition of ASPACE &TIMF~s(n),t(n)] to· be thefamily of languages accepted by an ASPACE[s(n)] machine with at(11) clock. Such a machine's accepting configuration is an acceptstate with th.e clock equal to 0, however the machine is neverallowed to look at its clock. With this definition Theorem 8.3tnakcs sense and is true for aU ten). It is now possible to ask,"What is ASPACE &TIME(s(n),t(n)] with t(n) > 2s(0) 1"

Corollary B.7:

a. ASPACE&TIM~llog(n),n*] = Var&Sz[*,n*] = Pl'IME

b. ASPACE &TIME[log(n),2n*] = Var &Sz[*,2n*] = PSPACE

proof: We have already shown line (a) in Corollary B.2. The lefthand equality of line (b) follows from Corollary B.4. lbe twocontainments of the second equality are proved as follows:

*(Var &S71*,2n 1C PSPACE): We are given an input structure, G,of size n, and a sentence, Fn ,with k variables and size 2nk. Checkwhether GI=Fn as follows:

Assumethat all -,'s have been pushed through to the inside andconsider the parse tree. for Fn. Each of the k variables may take onany of the n values of the universe of G. Starting at the leaves ofthe parse tree make a list of all the k-tuples of assignments whichmake the given nodes true in G. We can pass up the tree towardthe root computing the values making each node true as we go.

For example, an "&" node's list is gotten by intersecting thetwo lists it leads to, a "'Ix i " node gets those tuples <-,x2 ... xk>which are in the preceding node's list with all values of Xl •

When we reach the root either our list will have all nk

possibilities or it will be empty, since Fn has no .free variables.01=Fn if and only if we are in the former. case. At most· two ofthe k-tuples must be rcmbered at once, so PSPACE suffices.

* .(Var &Sz(*,2n J:> PSPACE): We show how to dcscnbe acomputation of Turing machine M runhing inIlSPACfo.:&TIME(s(n),t(n)]. We will construct fonnutas. Ct(p,x),meaning, "Symbol x occurs in cell P at time t." C

t(p,x) will be

written with O[log(s(n»/Iog(n)] variables.

rIlle idea is to say tbat there exists a triple of cell values X-I Xox1 til the previous move which lead to x in one move of M, and Xioccurs in r.cll p+ i at time t-1. In symbols:

I

= 3b1 \:I b2... Qs bs 31D1 ( IDo-+ b1 ... bs -+ IDl

& Accept(ID1] )

Accept[IDO]

ACCEPT21 (100 ) == 311)1 (EPath(Il)o ,ID1 ) & ACCEPT2a_l (101»

ACCEPT21-1 (lD1 ) == \:111)0 (APath(IDl ' IDo)~

ACCEPT2(a-1) (1°0 »

Note that in the above EPath(x,y) and APath(x,y) are theVar&Sz[O[s(n)/log(n)],s(n)2/log(n)] formulas which assert theexistence of a computation path from x to y .all of whoseintermediate states are existential, respectively universal. We canusc the abbreviation trick to conglomerate the two teoos on theright, making the size of Accepta (-) equal to:

aes(n).log(n) + Size(EPath)(a(n) + s(n) )s(n)/log(n), as desired.

The above corollary interested us especially because we nowhave natural classes, Log(CFL) and ASPACE&Att(togn,logn],identified with both of the the intermediate terms in the followingcontainment:

Ct (p,x) == 3x_1xOxl (X-1XoXl-+ 1-+ x & Ai=-l... l Ct-l(p+i~Xi»

We can usc the abbreviation trick to write Ct-l only once on theright. Note that p is a O[lo~(s(n»/log(n)]-tuple of variables coding

78

a tapc location Icss than 5(n). The sentence Cl

can be writtcn withO[log(s(n»/log(n)]- variablcs and size O[t(I1)]. For a polynotnials(n) this gives a constant nUlnber of variables as desired. I

Right now line (a) is what we call ASPACE[log(n)] and Var[*],but it is not clcar whether or not line (b) deserves at lcast t.he lattername.

Section C: Alternating Pebbling Games

In this section we present a new pebbling game to obtain lowelbounds for Var&Sz(w.o. Sue). This game is a modification ofEhrenfcllcht-Fraisse games. (See [Fra54] or [Ehr61].) Two playersplay the p pebble, m move game on a pair of structures G, H.Player I places pebbles on points from 0 or H trying todemonstrate a difference between theln while Player II matchesthese points trying to keep the structures looking the same. Wewill see in Theorem C.l that if Player II has a win for the p pebble,m move game on 0 and H, then G and H agree on all propertiesexpressible in Var&Sz(w.o. Suc)[p,m).

Definition: 'rhe.12 pcbble. In move game on G and H is defined asfollows: Initially the pebbles. gl ... gp • hI'" hp ,are off theboard. On move i, Player J picks lip a pebble gj (or hj ), 1~j~p,and places it on a vertex of G (or H). Player II answers by placingh. (or g.) on a corresponding point of H (or G). Let gj (i) be thep{>int o~ which gj is sitting just after move i. After each move i,O~i~m, define the map ~ as follows:

fi : cO ~ cH , gj (i) ~ hj (i)

The map fj takes the constants in 0 to the constants in H, andchosen points in G to the respective chosen points in H. We saythat Player II wins if for each i, O:S;i~m, fi is an isolnorphism ofthe induced substructures.

The quantifier rank of a sentence, cp, is the depth of .~csting ofquantifiers in cpo Since the quantifier rank of cp is obviously lessthan the size of cp, the following theoreln shows that the p,m gamegives a Var&Sz[p,m] lower bound on the expressibility of anyproperty on which G and H differ.

Theorem C.l: Player Il has a winning strategy for the p,m gameon G,H if and only if G and H agree on all sentences with pvariables and quantifier rank m.

We win give the proof, a minor Inodification of proofs in [Fra54]and [El1r61], shortly. First we will give an exmnple. Consider the 4pebble, d+ 1 move game on undirected graphs 0 and H where H isdisconnected while G is connected with diatneter d.

Player I wins the game as follows: On the first two moves hepute; pebbles h2 ,h] on vertices a,b such that a and b are indistinct components of H. Player II must place g2 ,g3 on somevertices e,f from G. There is a path of length at most d from e tof. Player I now uses the next d-l moves to walk along this pathwith pebbles ~ and 81. Player II must answer with a path in IIstarting at a, and thus never reaching b. Thus at move d + 1, twopebbles will coincide in G but not in H and Player I wins.

79

Figure I: 'I"he 4, d+ 1 game on G and H.

G

H

Notice that Player I's strategy was to follo~ the followingsentence, true in G but not in H: (Let M(u,v) == E(u,v) or u=v.)

Diam(d) == VX2VX 3 3) Xo (M(x2,xo) & 34 Xl (M(xo' xl) &

35 Xo(M(x1 ,xo)&...& 3d + 1 Xi( M(x1-i , Xi ) & M(xj , x3 » ...)

Also note that there is a sentence equivalent to I)jam(d) withonly 3 variables and log(d)+ 1 quantifier depth which Player Iwould have played had he known about it.

proof of Theorem C.l: (=»: Suppose there is a sentence S withp variables and quantifier rank m such that 0 satisfies S but Hdoes not. We must show that Player I wins the p pebble, m movegame on 0 and H. This is proved by induction on m:

base case: If m=O then G and H diffcr on a quantifier freesentencc, Le. the constants in 0 satisfy a fornlula that the constantsin H do not Thus they arc not isomorphic so Player I wins the 0move game.

inductive stcp: If S is of the form 'A, or A&B, then 0 and H

must disagree on one of A or B. Thus we may assume that S is ofthe fonn 3x1 M(x1 ). Here Player I places pebble g1 on somevertex gl (1) from G so that 01= M(g} (1». N~ Inatt~r what IIanswers we will have Ht= -lM(h1 (1». 11lus by Induction PlayerI will win.

Note that in the inductive step we have placed pebble g1 so wemust consider what happens if we later need gl again. The ,lflsweris that in tv1(gJ (l) ) the subst.itution of g) (1) is made for ,111 freeoccurrences of xI' If later on in the ganle we need to place g1again it will be for some sentcnce S' = QX1 N(x1)' Inside S' anoccurrence of xl are bound by QX 1 ' thus gl (1) does not occurand pebble gl may be safely reused.

( 4=): Conversely suppose that G and H agree on all sentenceswith p variables and quantifier rank m. We must show that PlayerII wins the p pebble, m move game on G and H. To facilitate aninductive proof wc Inust slightly strengthcn our claim. We provethe following:

Claim: Let k~p and stlpose that <O,clG ... ct

G) and <H, c1II ...

ckH) agree on al1 sentences S with new constants c1 .•. ct ' variablesxl'" xlr quantifier rank m, and such that nowhere in S does cioccur within the scope of a quantifier for xj ' Then Player II has awin for the p pebble, In Inove gatnc 011 G and H when started withthe first k pebbles on c1G ••. ctG and c1H ••. ct

Il respectively.

Notc that with k= 0 the claim reduces to what we need to show.

We prove the claim by induction on m. For m=O the mapfrom the constants in G to the constanle; in Hand c1G •.• ctG to

c111 ••• ckH must be an isomorphisln or else there would be a

quantifier free sentence on which the two structures disagree.

For the inductive step asume that the premise of the c1ailn holdsand let Player I n10ve placing, let us SHY, pebble gl on g1 (1).

Consider the (finite) collection of sentences SI (Xl) ... Sr (Xl ), in

the language of 0 together with constants c2 ••. ct such that <0,c2°... ckG) t= Si (gl (1». Let S == 3x1 ("i:1 ... r Si (Xl) ) .

Thus, <0,c20 ... ct o) t= S.

TI1US, by our assumption, <H, c2H •.• ctH> also S<ltisfics S. Let

h1(1) be a witness for Xl in S. Now <0,g1(1),c20 •.. ckG) and

<H,hl(1),c211 ••• ckH) agree on all sentences, R, of quantifier rank

m-l and variables xl'" lie such that no Ci occurs within the scopeof a quantifier for Xi' This is because any such R(cl ) satisfied byG would be an Si above and therefore also satisfied by .-1.

Our inductive asumption now shows that Player II wins theremaining 01-1 moves of the game, proving the claiR1- This proves

Theorem C.l . I

Before we apply Theorem C.l it is useful to give a slightlyditTerent characterization of G=varlt) H. What docs it mean whenG and H agree on all k variable sentences without successor? Theidea is that if Player ) chooses any r-tuple of points from G, r~k,

then there is a corresponding isomorphic r-tuple from H.Furthermore if ~Iayer I adds a point to the tuple in 0, and r<k,then there is a corresponding point in H which may be addedpreserving the isomorphism.

We have thus deduced the existence of a relation R on pairs ofr-tuples from G and r-tuples from H, Le. R ~ Ur=o...t ot X Hk ,satisfying:

a. R«>,(»

c. (R(g,h) & ,gl<k ) => (VxEG 3yEII R«g,x),<h,y»)

& (VyEU 3xEG R«g,x),<h,y»)

d. If g =<gl ... gr ), let ii = <g1 ... gj-l ,gi+ J ···gr) be ther-l tuple with gi rel1loved. 'Il,en:

R(g,h) => R(ii1ij) i= 1, ... , r

Proposition C.2: G=Var(k] H if and only if there exists a relation Rsatisfying (a - d) above.

proof: It should be clear that R corresponds to Player Irs winningstrategy in the k pebble ganlc Oil G and 1-1. 'nuls if such an Rexists then Player II can always win by lnatching chosen r-tuples inG with R-related r-tuples in If. Assunle R«gl (s) ... gt (s» ,<hl(s) ... hk (s») , Le. the chosen pOinl'i allcr 1110Ve s are R-related.Think of Player I's Inoving of pe~ble g.,i as two actions. I:'irst hepicks up gi' By (d) we know R(gi (S). -IIi (s»). Next he places gjback on some new point gj (s+ 1). By (c) there exists y in Hpreserving the relation, Le. with hi (s+1) =y, R(g(s+ 1),h(s .... I».In particular g(s+ 1) and h(s+ 1) are iSOlnorphic, and Player II wins.

Conversely, .if G=Varlk] H then define R from Player II'swinning strategy as follows:

R =

~'

rille k pebble gmne on 0 and H, started}«xl'" Xr)'<YI ... yr» with gi(O)=xi , hj(O) = Yi i= 1 ... r, is a

forced win for Player II.

The fact that Player II has a winning strategy for the k pebblegame on G and H gives us (a). (b), (c), and (d) follow fronl theru les of the game. I

Section I): Lower Bounds for Var(w.o. Suc)(kJ

Following [Fag76] and [BIHa79), we write certain axiolns forgraphs. First:

TO - Vx Vy ( .E(x,x) & [F~x,y) ~ F~y,x»))

To $Cays that G is loop free and undirected. We win assume inthis section that all graphs satisfy 10 .

~ix k and Jet 1~~k-l. The f(lJlowing sentences, St' , say thatfo~ any choice of di~tinct vertices, Xl ... Xi and xj + 1 ••• it-I' thereeXIsts a. vertex y different from the xi's with an edge to everyvertex In the first group and no edge to the second group.

Skj - 't/X1 ... VXt_1 ( ( "O<i<r<k Xi :I:. Xr ) =0>

3y [ "O<i<j+ 1 E(y,x i ) & "i<i<t (Y*X j & 'E(y,xi)'J)

We usc the Skj 's to write 1''1. ' an axiom which says that everyconceivable extension of a configuration of k-l points to aconfiguration of k points is reali1.able.

b. R(g,h) ~ g ,...., h

80

Theoreln III ([""38761, [BIHa79J): For any fixed k>O,

#O{ G 101==S, 101 =n} / #{ 01101 =n}

Thus the probability that none of a random n-(k-l) vertices is awitness for Skj is:

{ <i , j> I 1 ~ i ~ k, 1~ j ~ n }{«i1 ,jl >,<i2 ,j2 » I i1 ;t: i2 }=

=

Theorem 1).5: Clique(k+ I) is not in Var(w.o. Suc)[k].

proof: Recall that Clique(k + 1) is the set of graphs with acompletc subgraph of size k+ I. Clcarly any graph satisfying T. . C.· c( . ) k+1IS In Iqu k+l. Wc show that there exists a graph H 1= Tsuch that "k has no Ie +1 clique. Define the graph An = (~ ,E )as follows: n n

Theorem 1),4: "Hamilton Circuit" is not in Var(w.o. Suc}[*] .

Using similar techniques we can show the following:

Notice that An has no k+ 1 clique because any set of k+ 1vertices will have two with the same first coordinate.

Let An' = (Vn ,En' ) be a random subgraph of An ' Le. eachedge of En has probability 1/2 of being in E '. Now lim -+00

Prob(An' t= Tk ) = 1. (This follows from the ~me argumen~ as inthe proof of 'Illcorem 4.1, noting that every k-I tuple from V hasn points potentially satisfying Tk:) Let H be such a random

nA '.

Thus Hn satisfies Tk but has no k+1 clique~ fl.I

A counting argument shows that almost all graphs satisfy Tk .Define Pn (S), the probability that a graph of size n satisfies asentence S, as follows:

Iimn~OO Pn(1'k) =1

proof: Given j<k, and distinct vertices xl". xk-1 what is theprobability that a random vertex y is a witness for Sk.j? It's justthe probability that the k-] possible edges E(x i ,y) are correctly

present or absent, Le. l/2k-1 •

( 1- (l/2k-1 ) )n-(k-l)

The probability that any of the fewer than nk sequences, xl •••

xk-l ,. j , cause Tk to fail is less than

nk • at

and this last probability goes to 0 as n goes to infinity.

We are interested in Tk because of the next result:

Theoreln D.2: For any two graphs 0 and H,

(Ot=Tk&Ht=Tk ) => O=var[k]H

proof: Tk says that every k-l tuple may be extended to a k tuplein any conceivable way. It follows that the relation:

R = {«a1 ... ar >,<b1 ... br »J O~r~k, ai EO, bi EH, & }<a1 ... ak>~ <b1 ..• bk >

Almost a11 graphs have a Hamilton circuit; however, in [811-Ia79]it is shown that for any k there is a graph Hk which satisfies Tkand yet has no Hanli1ton circuit. Jt follows that there exist twographs, Ok' Hk , both satisfying l'k and yet differing on theproperty of having a Hamilton circuit. Thus:

satisfies (a) - (d) of Proposition 4.3. Therefore 0 =Var[k] H. I

Corollary 0.,3: Oraph Isomorphism is not in Var(w.o. Suc)[k] .

proof: (f OraphIso were in Var(w.o. Suc)[k] then there would besentences F1 ,F2 ... with k variables each such that for graphs Gand H of size n,

By Theorem 4.1 there exist two non-isomorphic graphs Ok andlit both satisfying Tt . Clearly (Ok' G~?t=Fn. But by Theorem4.2, Ok ===Var[k] Hk · It follows that Player II wins the k pebblegame on <Ok' Hk>and <Ok' Ok>. Her strategy is to answer pointsin the first compof.'..mt with the same point in the other copy of Ok'and to use Player II's winning strategy for the k pebble game onOk and Hk to answer moves in the right component Thus,

<O,H> t= Fn

(Ok ,Hk > 1= Fn ,but Ok

This contradiction proves the corollary.

G~H

is not isolnorphic to Hk •

I

Section D: Conclusions

We have shown that first order expressibility is a viable view ofcomputational complexity. We fcel that it is a .natural way to

obtain both· upper and lower bounds. lbe alternating pebblinggames make the finding of optimal descriptions of graph properties(without successor) a tractable problem. Furthermore oursimulation theorems show that optimal sentences (with successor)for a property C can be easily translated to nearly optimalalgorithms for checking C.

The following general areas of exploration are suggested:

(1): Find upper and lower bounds on Var&Sz(w.o. Sue) for acollection of graph problems such as planarity, graphhomeomorphism, vertex matching, ctc.

(2): Improve the simulations of Section n, and then try to proveoptimality. Exactly how many variables are needed todescribe a DTIMF~nkJ computation?

(3): Develop techniques to prove lower bounds on Var&Sz, i.e.with successor. This seems hard but worthwhile; possibletechniques are discussed in [hn80a] and [lm80b).

81

Acknowledgements

Wann thanks to Juris Jlartmanis, my thesis adviser. Manythanks to John Hopcroft, Albert Meyer, and Michael· Morley forhelpful technical discussions. Thanks to MITs Laboratory forComputer Science for letting me visit the summer of 1980, whereand when this paper was completed.

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